\(\int \frac {\cos ^2(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\) [848]

Optimal result

Integrand size = 42, antiderivative size = 427 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {\left (a^2 B-3 b^2 B+2 a b C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^2 b \sqrt {a+b} d}+\frac {(3 b B+a (B-2 C)) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^2 \sqrt {a+b} d}+\frac {\sqrt {a+b} (3 b B-2 a C) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^3 d}+\frac {B \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}+\frac {b \left (a^2 B-3 b^2 B+2 a b C\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \] Output:

(B*a^2-3*B*b^2+2*C*a*b)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^ 
(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c 
))/(a-b))^(1/2)/a^2/b/(a+b)^(1/2)/d+(3*B*b+a*(B-2*C))*cot(d*x+c)*EllipticF 
((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c)) 
/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^2/(a+b)^(1/2)/d+(a+b)^(1/2 
)*(3*B*b-2*C*a)*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),( 
a+b)/a,((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+ 
c))/(a-b))^(1/2)/a^3/d+B*sin(d*x+c)/a/d/(a+b*sec(d*x+c))^(1/2)+b*(B*a^2-3* 
B*b^2+2*C*a*b)*tan(d*x+c)/a^2/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1597\) vs. \(2(427)=854\).

Time = 16.19 (sec) , antiderivative size = 1597, normalized size of antiderivative = 3.74 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

Integrate[(Cos[c + d*x]^2*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[ 
c + d*x])^(3/2),x]
 

Output:

((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*((-2*b*(b*B - a*C)*Sin[c + d*x])/(a 
^2*(-a^2 + b^2)) + (2*(-(b^3*B*Sin[c + d*x]) + a*b^2*C*Sin[c + d*x]))/(a^2 
*(a^2 - b^2)*(b + a*Cos[c + d*x]))))/(d*(a + b*Sec[c + d*x])^(3/2)) - ((b 
+ a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(3/2)*Sqrt[(1 - Tan[(c + d*x)/2]^2)^( 
-1)]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[( 
c + d*x)/2]^2)]*(a^3*B*Tan[(c + d*x)/2] + a^2*b*B*Tan[(c + d*x)/2] - 3*a*b 
^2*B*Tan[(c + d*x)/2] - 3*b^3*B*Tan[(c + d*x)/2] + 2*a^2*b*C*Tan[(c + d*x) 
/2] + 2*a*b^2*C*Tan[(c + d*x)/2] - 2*a^3*B*Tan[(c + d*x)/2]^3 + 6*a*b^2*B* 
Tan[(c + d*x)/2]^3 - 4*a^2*b*C*Tan[(c + d*x)/2]^3 + a^3*B*Tan[(c + d*x)/2] 
^5 - a^2*b*B*Tan[(c + d*x)/2]^5 - 3*a*b^2*B*Tan[(c + d*x)/2]^5 + 3*b^3*B*T 
an[(c + d*x)/2]^5 + 2*a^2*b*C*Tan[(c + d*x)/2]^5 - 2*a*b^2*C*Tan[(c + d*x) 
/2]^5 - 6*a^2*b*B*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b) 
]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[ 
(c + d*x)/2]^2)/(a + b)] + 6*b^3*B*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]] 
, (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d 
*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 4*a^3*C*EllipticPi[-1, ArcSin[ 
Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + 
 b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - 4*a*b^2*C*Ell 
ipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d 
*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a...
 

Rubi [A] (verified)

Time = 2.00 (sec) , antiderivative size = 468, normalized size of antiderivative = 1.10, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4560, 3042, 4522, 27, 3042, 4549, 27, 3042, 4546, 3042, 4409, 3042, 4271, 4319, 4492}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 4560

\(\displaystyle \int \frac {\cos (c+d x) (B+C \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {B+C \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 4522

\(\displaystyle \frac {B \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {-b B \sec ^2(c+d x)+3 b B-2 a C}{2 (a+b \sec (c+d x))^{3/2}}dx}{a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {B \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {-b B \sec ^2(c+d x)+3 b B-2 a C}{(a+b \sec (c+d x))^{3/2}}dx}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {B \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {-b B \csc \left (c+d x+\frac {\pi }{2}\right )^2+3 b B-2 a C}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{2 a}\)

\(\Big \downarrow \) 4549

\(\displaystyle \frac {B \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {-\frac {2 \int -\frac {b \left (B a^2+2 b C a-3 b^2 B\right ) \sec ^2(c+d x)-2 a b (b B-a C) \sec (c+d x)+\left (a^2-b^2\right ) (3 b B-2 a C)}{2 \sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (a^2 B+2 a b C-3 b^2 B\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {B \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\int \frac {b \left (B a^2+2 b C a-3 b^2 B\right ) \sec ^2(c+d x)-2 a b (b B-a C) \sec (c+d x)+\left (a^2-b^2\right ) (3 b B-2 a C)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (a^2 B+2 a b C-3 b^2 B\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {B \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\int \frac {b \left (B a^2+2 b C a-3 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-2 a b (b B-a C) \csc \left (c+d x+\frac {\pi }{2}\right )+\left (a^2-b^2\right ) (3 b B-2 a C)}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (a^2 B+2 a b C-3 b^2 B\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}\)

\(\Big \downarrow \) 4546

\(\displaystyle \frac {B \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {b \left (a^2 B+2 a b C-3 b^2 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+\int \frac {\left (a^2-b^2\right ) (3 b B-2 a C)+\left (-2 a b (b B-a C)-b \left (B a^2+2 b C a-3 b^2 B\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (a^2 B+2 a b C-3 b^2 B\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {B \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {b \left (a^2 B+2 a b C-3 b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {\left (a^2-b^2\right ) (3 b B-2 a C)+\left (-2 a b (b B-a C)-b \left (B a^2+2 b C a-3 b^2 B\right )\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (a^2 B+2 a b C-3 b^2 B\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}\)

\(\Big \downarrow \) 4409

\(\displaystyle \frac {B \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {b \left (a^2 B+2 a b C-3 b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (a^2-b^2\right ) (3 b B-2 a C) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx-b (a-b) (a (B-2 C)+3 b B) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (a^2 B+2 a b C-3 b^2 B\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {B \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\left (a^2-b^2\right ) (3 b B-2 a C) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (a^2 B+2 a b C-3 b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (a-b) (a (B-2 C)+3 b B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (a^2 B+2 a b C-3 b^2 B\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}\)

\(\Big \downarrow \) 4271

\(\displaystyle \frac {B \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {b \left (a^2 B+2 a b C-3 b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (a-b) (a (B-2 C)+3 b B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) (3 b B-2 a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{a \left (a^2-b^2\right )}-\frac {2 b \left (a^2 B+2 a b C-3 b^2 B\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {B \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {b \left (a^2 B+2 a b C-3 b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) (3 b B-2 a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}-\frac {2 (a-b) \sqrt {a+b} (a (B-2 C)+3 b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{a \left (a^2-b^2\right )}-\frac {2 b \left (a^2 B+2 a b C-3 b^2 B\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {B \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {-\frac {2 (a-b) \sqrt {a+b} \left (a^2 B+2 a b C-3 b^2 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}-\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) (3 b B-2 a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}-\frac {2 (a-b) \sqrt {a+b} (a (B-2 C)+3 b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{a \left (a^2-b^2\right )}-\frac {2 b \left (a^2 B+2 a b C-3 b^2 B\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}\)

Input:

Int[(Cos[c + d*x]^2*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d* 
x])^(3/2),x]
 

Output:

(B*Sin[c + d*x])/(a*d*Sqrt[a + b*Sec[c + d*x]]) - (((-2*(a - b)*Sqrt[a + b 
]*(a^2*B - 3*b^2*B + 2*a*b*C)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec 
[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + 
 b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d) - (2*(a - b)*Sqrt[a + 
b]*(3*b*B + a*(B - 2*C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + 
d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]* 
Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/d - (2*Sqrt[a + b]*(a^2 - b^2)*(3 
*b*B - 2*a*C)*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[c + 
 d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)] 
*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(a*d))/(a*(a^2 - b^2)) - (2*b*(a 
^2*B - 3*b^2*B + 2*a*b*C)*Tan[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[a + b*Sec[c 
+ d*x]]))/(2*a)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4271

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a 
 + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) 
*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ 
c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && 
NeQ[a^2 - b^2, 0]
 

rule 4319

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* 
x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt 
[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, 
 f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4409

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ 
.) + (a_)], x_Symbol] :> Simp[c   Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + 
Simp[d   Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, 
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
 

rule 4492

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c 
sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a 
 + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e 
+ f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + 
 f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, 
 f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
 

rule 4522

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e 
 + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(a*f*n)), x] + Sim 
p[1/(a*d*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*B* 
n - A*b*(m + n + 1) + A*a*(n + 1)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f* 
x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] 
 && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
 

rule 4546

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Int[(A + (B - C 
)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Simp[C   Int[Csc[e + f*x]*(( 
1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A 
, B, C}, x] && NeQ[a^2 - b^2, 0]
 

rule 4549

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_. 
) + (a_))^(m_), x_Symbol] :> Simp[(A*b^2 + a^2*C)*Cot[e + f*x]*((a + b*Csc[ 
e + f*x])^(m + 1)/(a*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(a*(m + 1)*(a^2 - 
 b^2))   Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*(a^2 - b^2)*(m + 1) - a*b* 
(A + C)*(m + 1)*Csc[e + f*x] + (A*b^2 + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], 
x], x] /; FreeQ[{a, b, e, f, A, C}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m 
] && LtQ[m, -1]
 

rule 4560

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. 
)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) 
*(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2   Int[(a + b*Csc[e + f*x])^(m 
+ 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ 
[{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1610\) vs. \(2(396)=792\).

Time = 12.50 (sec) , antiderivative size = 1611, normalized size of antiderivative = 3.77

method	result	size
default	\(\text {Expression too large to display}\)	\(1611\)

Input:

int(cos(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x,me 
thod=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

1/d/a^2/(a+b)/(a-b)*((6*cos(d*x+c)^2+12*cos(d*x+c)+6)*B*(1/(a+b)*(b+a*cos( 
d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b*Elli 
pticPi(-csc(d*x+c)+cot(d*x+c),-1,((a-b)/(a+b))^(1/2))+(-6*cos(d*x+c)^2-12* 
cos(d*x+c)-6)*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c 
)/(cos(d*x+c)+1))^(1/2)*b^3*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,((a-b)/(a 
+b))^(1/2))+(-4*cos(d*x+c)^2-8*cos(d*x+c)-4)*C*(1/(a+b)*(b+a*cos(d*x+c))/( 
cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^3*EllipticPi(-csc 
(d*x+c)+cot(d*x+c),-1,((a-b)/(a+b))^(1/2))+(4*cos(d*x+c)^2+8*cos(d*x+c)+4) 
*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c) 
+1))^(1/2)*a*b^2*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,((a-b)/(a+b))^(1/2)) 
+(-cos(d*x+c)^2-2*cos(d*x+c)-1)*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1) 
)^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^3*EllipticE(-csc(d*x+c)+cot(d* 
x+c),((a-b)/(a+b))^(1/2))+(-cos(d*x+c)^2-2*cos(d*x+c)-1)*B*(1/(a+b)*(b+a*c 
os(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b*E 
llipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(3*cos(d*x+c)^2+6*cos 
(d*x+c)+3)*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/( 
cos(d*x+c)+1))^(1/2)*a*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^ 
(1/2))+(3*cos(d*x+c)^2+6*cos(d*x+c)+3)*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d* 
x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^3*EllipticE(-csc(d*x+c) 
+cot(d*x+c),((a-b)/(a+b))^(1/2))+(-2*cos(d*x+c)^2-4*cos(d*x+c)-2)*C*(1/...
 

Fricas [F]

\[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \cos \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(cos(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2 
),x, algorithm="fricas")
 

Output:

integral((C*cos(d*x + c)^2*sec(d*x + c)^2 + B*cos(d*x + c)^2*sec(d*x + c)) 
*sqrt(b*sec(d*x + c) + a)/(b^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), 
 x)
 

Sympy [F]

\[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(cos(d*x+c)**2*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**( 
3/2),x)
 

Output:

Integral((B + C*sec(c + d*x))*cos(c + d*x)**2*sec(c + d*x)/(a + b*sec(c + 
d*x))**(3/2), x)
 

Maxima [F]

\[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \cos \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(cos(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2 
),x, algorithm="maxima")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*cos(d*x + c)^2/(b*sec(d*x + 
c) + a)^(3/2), x)
 

Giac [F]

\[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \cos \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(cos(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2 
),x, algorithm="giac")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*cos(d*x + c)^2/(b*sec(d*x + 
c) + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:

int((cos(c + d*x)^2*(B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/cos(c + d* 
x))^(3/2),x)
 

Output:

int((cos(c + d*x)^2*(B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/cos(c + d* 
x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{2} b^{2}+2 \sec \left (d x +c \right ) a b +a^{2}}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )}{\sec \left (d x +c \right )^{2} b^{2}+2 \sec \left (d x +c \right ) a b +a^{2}}d x \right ) b \] Input:

int(cos(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x)
 

Output:

int((sqrt(sec(c + d*x)*b + a)*cos(c + d*x)**2*sec(c + d*x)**2)/(sec(c + d* 
x)**2*b**2 + 2*sec(c + d*x)*a*b + a**2),x)*c + int((sqrt(sec(c + d*x)*b + 
a)*cos(c + d*x)**2*sec(c + d*x))/(sec(c + d*x)**2*b**2 + 2*sec(c + d*x)*a* 
b + a**2),x)*b